Question

If $\sin x + \cos x + \tan x + \cot x + \sec x + \cos ecx = 7$ and $\sin 2x = a - b\sqrt 7$, then ordered pairs $(a,b)$ is,
A. $(6,2)$
B. $(8,3)$
C. $(22,8)$
D. $(11,4)$

Answer 1

To solve the problem we have to first convert the first given condition into only $\sin$and $\cos$terms, as in condition two, there will be only $\sin$and $\cos$terms. Then we will have to perform required operations to convert the first condition into a form like the second condition in order to get the answer.

Complete step by step answer:
The first condition is:
$\sin x + \cos x + \tan x + \cot x + \sec x + \cos ecx = 7$
Now, converting $\tan x,\cot x,\sec x,\cos ecx$ into $\sin x$and $\cos x$terms. We know that secant and cosecant functions are reciprocals functions of cosine and sine. Also,
$\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$
By converting them, we get,
$\Rightarrow \sin x + \cos x + \left( {\dfrac{{\sin x}}{{\cos x}}} \right) + \left( {\dfrac{{\cos x}}{{\sin x}}} \right) + \left( {\dfrac{1}{{\cos x}}} \right) + \left( {\dfrac{1}{{\sin x}}} \right) = 7$
$\Rightarrow (\sin x + \cos x) + \left( {\dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\sin x\cos x}}} \right) + \left( {\dfrac{{\sin x + \cos x}}{{\sin x\cos x}}} \right) = 7$

We know the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$, so, substituting this value in the equation, we get,
$\Rightarrow (\sin x + \cos x) + \left( {\dfrac{1}{{\sin x\cos x}}} \right) + \left( {\dfrac{{\sin x + \cos x}}{{\sin x\cos x}}} \right) = 7$
Taking the term $\left( {\dfrac{1}{{\sin x\cos x}}} \right)$ on the Right Hand Side, we get,
$\Rightarrow (\sin x + \cos x) + \left( {\dfrac{{\sin x + \cos x}}{{\sin x\cos x}}} \right) = 7 - \left( {\dfrac{1}{{\sin x\cos x}}} \right)$
Taking $(\sin x + \cos x)$ common in Left Hand Side, we get,
$\Rightarrow \left( {\sin x + \cos x} \right)\left( {1 + \dfrac{1}{{\sin x\cos x}}} \right) = 7 - \left( {\dfrac{1}{{\sin x\cos x}}} \right)$

Now, we know the trigonometric formula $\sin 2x = 2\sin x\cos x$. So, we get,
$\Rightarrow \left( {\sin x + \cos x} \right)\left( {1 + \dfrac{2}{{\sin 2x}}} \right) = 7 - \left( {\dfrac{2}{{\sin 2x}}} \right)$
Now, squaring both sides, we get,
$\Rightarrow {\left( {\sin x + \cos x} \right)^2}{\left( {1 + \dfrac{2}{{\sin 2x}}} \right)^2} = {\left[ {7 - \left( {\dfrac{2}{{\sin 2x}}} \right)} \right]^2}$
Using algebraic identities for simplification of the expression, we get,
$\Rightarrow ({\sin ^2}x + {\cos ^2}x + 2\sin x\cos x)\left( {1 + \dfrac{4}{{\sin 2x}} + \dfrac{4}{{{{\sin }^2}2x}}} \right) = 49 - \dfrac{{28}}{{\sin 2x}} + \dfrac{4}{{{{\sin }^2}2x}}$
$\Rightarrow (1 + \sin 2x)\left( {1 + \dfrac{4}{{\sin 2x}} + \dfrac{4}{{{{\sin }^2}2x}}} \right) = 49 - \dfrac{{28}}{{\sin 2x}} + \dfrac{4}{{{{\sin }^2}2x}}$
[Using, ${\sin ^2}x + {\cos ^2}x = 1$ and $\sin 2x = 2\sin x\cos x$]

Now, opening the brackets, we get,
$\Rightarrow \left( {1 + \dfrac{4}{{\sin 2x}} + \dfrac{4}{{{{\sin }^2}2x}}} \right) + \sin 2x\left( {1 + \dfrac{4}{{\sin 2x}} + \dfrac{4}{{{{\sin }^2}2x}}} \right) = 49 - \dfrac{{28}}{{\sin 2x}} + \dfrac{4}{{{{\sin }^2}2x}}$
$\Rightarrow 1 + \dfrac{4}{{\sin 2x}} + \dfrac{4}{{{{\sin }^2}2x}} + \sin 2x + 4 + \dfrac{4}{{\sin 2x}} = 49 - \dfrac{{28}}{{\sin 2x}} + \dfrac{4}{{{{\sin }^2}2x}}$
Now, on simplifying the equation, we get,
$\Rightarrow \dfrac{8}{{\sin 2x}} + \sin 2x + 5 = 49 - \dfrac{{28}}{{\sin 2x}}$
Making Right Hand Side $0$ by changing the side of all terms to Left Hand Side, we get,
$\Rightarrow \dfrac{8}{{\sin 2x}} + \sin 2x + 5 - 49 + \dfrac{{28}}{{\sin 2x}} = 0$
\Rightarrow \dfrac{{36}}{{\sin 2x}} + \sin 2x + 44 = 0 \\\
Multiplying both sides by $\sin 2x$, we get,
$\Rightarrow {\sin ^2}2x + 44\sin 2x + 36 = 0$

Now, using Quadratic formula, to solve the above quadratic equation, we get,
\Rightarrow \sin 2x = \dfrac{{ - 44 \pm \sqrt {{{(44)}^2} - 4\left( 1 \right)\left( {36} \right)}\sin 2x = a - b\sqrt 7 }}{{2\left( 1 \right)}}
$\Rightarrow \sin 2x = \dfrac{{ - 44 \pm \sqrt {1936 - 144} }}{{2\left( 1 \right)}}$
$\Rightarrow \sin 2x = \dfrac{{ - 44 \pm \sqrt {1792} }}{2}$
$\Rightarrow \sin 2x = \dfrac{{ - 44 \pm \sqrt {256 \times 7} }}{2}$
$\Rightarrow \sin 2x = \dfrac{{ - 44 \pm 16\sqrt 7 }}{2}$
$\Rightarrow \sin 2x = - 22 \pm 8\sqrt 7 - - - - \left( 1 \right)$
Now, the given value .
Comparing the condition to equation $(1)$, we get,
$a = 22,b = 8$
Therefore, the ordered pair $(a,b)$ is $(22,8)$.

Hence, the correct answer is option C.

Note: The above problem can also be solved by not changing $\sin x\cos x$ into $\sin 2x$. Instead we can take $\sin x\cos x = p$ and perform the same functions as above and get the result of the quadratic equation. The second condition becomes $2p$, so we can adjust the values as required. But, the ultimate result will be $(22,8)$.